Problem: $g(x)=\dfrac{8x-1}{x+4}$ $h(x)=3x+10$ Write $(g\circ h)(x)$ as an expression in terms of $x$. $(g\circ h)(x)=$
First, let's write $(g\circ h)(x)$ as $g(h(x))$. Next, we write $h(x)$ as the input to function $g$. $g({h(x)})=\dfrac{8({h(x)})-1}{({h(x)})+4}$ Since $h(x)=3x+10$, this becomes: $\begin{aligned} g({h(x)})&=\dfrac{8({3x+10})-1}{({3x+10})+4}\\ \\ &=\dfrac{24x+80-1}{3x+10+4}\\ \\ &=\dfrac{24x+79}{3x+14}\\ \\ \end{aligned}$ Note: We simplified the result to obtain a nicer expression, but this is not necessary. The answer: $(g\circ h)(x)=\dfrac{24x+79}{3x+14}$